Chapter 9

Systems of Particles

27/15/04

Center of Mass

How do I define the position of the cow?What is h for U = mgh?

Take average position of mass: “Center of mass”Treat extended objects as point particles at CM

Real objects are not point particles

(Easy for spherical objects)

37/15/04

Center of Mass (Discrete)

To find the center of mass for a bunch of point masses:

m1, m2, m3,… at r1, r2, r3,…

rCM

r1

r2

r3

iii

total

totalcm

rmM

M

rmrmr

1

...2211

47/15/04

For an extended object, break it into point masses (mass = dm)

r

Center of Mass (Continuous)

dmrM

rtotal

cm 1

Note: This equation assumes that density is constant

57/15/04

Center of Mass (Demo)

021

2211

mm

rmrmrcm

m1

0.5 kg

4 3

m2

???

To balance, rcm = 0

kgm

kg

r

rmm

rmmr

67.0

3

)4)(5.0(

0

2

2

112

2211

67/15/04

Center of Mass (Bigger Demo)

m1

14.7 kg

141

m2

???

Again, to balance: rcm = 0

021

2211

mm

rmrmrcm

kgm

kg

r

rmm

05.1

14

)1)(7.14(

2

2

112

As before: Why didn’t it work this time?!

77/15/04

Center of Mass (Bigger Demo)

m1

14.7 kg

141

m2

???

We forgot about the bar!

mbar=0.745 kg

7

CM in center of bar:

kgkgmbar 695.0)745.0(15

142,

0.5kgkgmbar 050.0)745.0(

15

11,

87/15/04

Center of Mass (Bigger Demo)

m1

14.7 kg

141

m2

???

021

2,2,1,1,2211

bar

barbarbarbarcm mmm

rmrmrmrmr

kgkgkgkg

m

r

rmrmrmm barbarbarbar

015.114

)7)(695.0()5.0)(050.0()1)(7.14(2

2

2,2,1,1,112

97/15/04

Center of Mass Example

m1 = 3 kg

m2 = 5 kg

m1 at position (1,1) meterm2 at position (2,0) meter

Mtotal = 8 kgrCM

ji

kg

ikgjikg

M

rmrmr

totalcm

ˆ8

10ˆ8

13

8

)ˆ2)(5()ˆˆ)(3(2211

107/15/04

Center of Mass for 2D objects

Often don’t need to do integral: Use symmetry

Can also add symmetric objects together

m2 = 5 kg

m1 = 3 kg

rCM

117/15/04

Another Way to Determine CM

127/15/04

Why Do We Want Center of Mass?

Can treat extended objects or groups of objects as points

Gravity pulls on CM: h = hCM for Ugrav

aCM = Ftot /Mtot

Brick falls if CM not supported

137/15/04

Mystery Hill

Gravity pulls down on the CM of the system…

147/15/04

Example: (Problem 9.9)A cylindrical can with mass M, height H, and uniform density is initially filled with soda/pop of mass m. We then punch a hole in the top and bottom so that the liquid can drain out. Find:

a) The center of mass of the can when full

b) The center of mass of the can when empty

c) What happens to the height of the CM while draining?

d) Find the height of the soda/pop when the CM reaches its lowest point

This one is pretty tough!

157/15/04

Ftot = dp/dt

Momentum:

New fundamental quantity (like force,energy,..)

For point particle p = mv

For extended object: pCM = mvCM

For group of particles: ptotal=m1v1+m2v2+...

Relation to Force: Ftot = ma

= m dv/dt

= d[mv]/dt

167/15/04

Conservation of Momentum

If Ftot = 0, then momentum is constant

For an isolated system (no external forces):

Ftot = dp/dt

pinitial = pfinal

True also for groups of particles:

If Fexternal = 0, then pCM = constant

Even if there are internal forces

177/15/04

Momentum of a Noisy Cricket

Will Smith: 200 lbs ~91 kg

Thrown back ~3m in 1s 3 m/s

smkgsmkg

vmP SmithSmithSmith

/273)/3)(91(

PSmith PPulse

PulseSmith

PulseSmith

PP

PP

0

smkgPPulse /273

187/15/04

Momentum of a Noisy Cricket

Looking ahead to Phys 214:

J

smsmkg

pcE

10

8

102.8

)/103)(/273(

So how much energy is this?

m

p

m

mvmvK

22

)(

2

1 222

But light is massless…

Recall: A car at 60 mph has ~ 105 J!

197/15/04

The Celebrated Jumping Frog of Calaveras County

A 2 kg frog sits on the end of a motionless, floating 50 kg log (3 meters long).

3 m

She jumps to the other end.

How far does the log move due to the jump?

207/15/04

Celebrated Jumping Frog (cont.)

Consider frog and log together as a system of objects

Initial:

Final:x

No external forces momentum of system doesn’t change

pinitial = pfinal = 0

Center of mass of system does not move

217/15/04

xx = 0 x = 1.5m x = 3m

xCM = 1.53 m

1.53 m

initial

final

1m

CM does not move, log moves to keep it in same place!

1.53 m

From geometry, we see that:

Celebrated Jumping Frog (cont.)

mkg

mkgmkg

mm

xmxmx

logfrog

loglogfrogfrogcm

53.151

)5.1)(50()3)(1(

xmx

mxmm

06.0

353.153.1

227/15/04

Gun and Bullet

Both are at rest when fired, no net external force

Total Momentum: pinitial = pfinal = 0

What does this say about the motion after the gunis fired?

vG vB

237/15/04

Gun and Bullet (continued)

mB << mG so vG << vB

The gun is heavier so it moves slower!

vG vB

0 finalinitial pp

0 ggbbfinal vmvmp

bg

bg v

m

mv

247/15/04

Gun and Bullet (Question 1)

Does the Center of Mass of the system move?

No, because pinitial = pfinal = 0 for the system!gg

ggbb

gg

ggbbcmcm

mm

vmvm

mm

rmrm

dt

d

dt

rdv

vG vB

257/15/04

Gun and Bullet (Question 2)

If the momenta are the same, why do you want to hold the gun rather than catch the bullet?

Consider the kinetic energy:

K = ½mv2 = p2/(2m)

|pG| = |pB| and mB << mG KG << KB

vG vB

267/15/04

Spaceship Example

A spaceship in outer space, free from any interactions with other objects, has mship=34,000 kg and vship,i=1,000 m/s i. The ship launches a satellite msat=1,000 kg which moves with vsat,f=(1,500m/s)i + (400 m/s)j. What is the velocity of the spaceship after the satellite launch?

277/15/04

Spaceship (continued)

No external forces, so momentum is conserved:

fsatsatfshipshipishipsatship

fi

vmvmvm

pp

,,,

Solve for vship,f

ship

fsatsatishipsatshipfship m

vmvmv ,,

,

jsmismv

kg

jsmismkgismkgv

fship

fship

/8.11/985

000,34

ˆ)/400(ˆ)/500,1()000,1()ˆ/000,1)(000,35(

,

,

287/15/04

Spaceship (question)

Is Kinetic Energy conserved?

Kinitial = Kship = ½mtotal(vinitial)2

= ½(3.5 x 104 kg)(1000 m/s)2 = 1.75 x 1010 J

Kfinal = Kship + Ksatellite = ½mshipvship2 + ½msatvsat

2

= ½(3.4 x 104 kg)[(985 m/s)2 + (11.8 m/s)2] + ½(1000 kg)[(1500 m/s)2 + (400 m/s)2]

= 1.77 x 1010 J

NO! Extra energy added when satellite ejected.

297/15/04

Rockets

A gun shooting a bullet is a discrete process

We can generalize this to continuous processes

Rockets operate by shooting outa continuous stream of gas

Note: Rocket propulsion was

originally thought to be impossible…Nothing to push against!!

307/15/04

Rockets

= MvPi

mvvM Mv

v m

mvmvvMmv = Mv

v-vmvv+mM- =

P = PP

ex

ex

ex

fuelrocketf

)())((

exmvv=M

317/15/04

Rockets

exvmv=M

dMvM dv= ex

dmv=dvM ex

Δm 0

Note: dm = -dM

Dividing both sides by dt dt

dMv=

dt

dvM ex

aMRvex Let dM/dt = -R, where R is the rate of fuel consumption

327/15/04

Rockets

dMM

v dv= ex

f

i

f

i

M

M

ex

v

v M

dM = -vdv

f

iexif M

M+v=vv ln

To find the velocity of the rocket we must integrate

Note: Velocity is a function of mass loss, not time